Question

8. What are the zeroes of the polynomial function f(x) =
x2(2x + 1)(x - 1)? with solution​

Answer 1

Answer:

x2(2x+1)(x-1)

x2(3x-1x+1)(x-1)

x(x+1)+3(x-1x)(x-1)

(x+3)(x+1)(x-1x)(x-1)

(2x-4)(2x+2x)

(-2x)(4x)

x=-2;x=4

Answer 2

Given,

$\[f\left( x \right) = x2\left( {2x + 1} \right)\left( {x - 1} \right)\]$

Solution,

Know that zeros of a polynomial are the values of x for which the function has a value equal to zero.

Therefore,

$\[\begin{array}{l} \Rightarrow x2\left( {2x + 1} \right)\left( {x - 1} \right) = 0\\ \Rightarrow x\left( {4x + 2} \right)\left( {x - 1} \right) = 0\\ \Rightarrow x\left( {4{x^2} - 4x + 2x - 2} \right) = 0\end{array}\]$

Solve further,

$\[\begin{array}{l} \Rightarrow x\left( {4{x^2} - 2x - 2} \right) = 0\\ \Rightarrow 4{x^3} - 2{x^2} - 2x = 0\\ \Rightarrow x = 1,0,\frac{1}{2}\end{array}\]$

Hence, the value of zeros are $\[1,0\,{\rm{and}}\,\frac{1}{2}.\]$